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Find the limit $$\lim\limits_{x\to\infty}\int^{2x}_{x}\frac{1}{t}dt$$

My trial

\begin{align}\lim\limits_{x\to\infty}\int^{2x}_{x}\frac{1}{t}dt &= \lim\limits_{x\to\infty}\large\left[\ln t \large\right]^{2x}_{x}\\&= \lim\limits_{x\to\infty}\large\left[\ln 2x -\ln x \large\right]\end{align} This yields the indeterminate form $\infty-\infty.$ I'm thinking of applying L'Hopital's rule but no headway. Any hints, please?

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    $\begingroup$ Hint: $\ln a-\ln b=\cdots\,$? $\endgroup$ – David Mar 26 at 5:58
  • $\begingroup$ @David: That's $\ln(\frac{a}{b})$ $\endgroup$ – Omojola Micheal Mar 26 at 6:00
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    $\begingroup$ Thanks everyone! Was having a wrong thought, initially! I didn't think through! $\endgroup$ – Omojola Micheal Mar 26 at 6:03
  • $\begingroup$ Isn't there a way I could apply L'Hopital's rule without having to evaluate the limit directly? @Everyone. I am really eager to apply this to L'Hopital's rule. $\endgroup$ – Omojola Micheal Mar 26 at 6:30
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$\displaystyle \lim_{x \to \infty} \int_x^{2x} \dfrac{dt}{t} = \lim_{x \to \infty} [\ln(2x) - \ln x] = \lim_{x \to \infty}[\ln x + \ln 2 - \ln x] = \lim_{x \to \infty} [\ln 2] = \ln 2. \tag 1$

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    $\begingroup$ Isn't there a way I could apply L'Hopital's rule without having to evaluate the limit directly? I am really eager to apply this to L'Hopital's rule. $\endgroup$ – Omojola Micheal Mar 26 at 6:35
  • $\begingroup$ @OmojolaMicheal: can't see anything off hand; that doesn't mean it's not possible! $\endgroup$ – Robert Lewis Mar 26 at 6:41
  • $\begingroup$ Okay, thanks! I appreciate! $\endgroup$ – Omojola Micheal Mar 26 at 6:47
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You're almost there!

$$\lim_{x\to\infty}(\ln 2x-\ln x)=\lim_{x\to\infty}\ln\left(\frac{2x}{x}\right)=\lim_{x\to\infty}\ln 2=\ln 2$$

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Your evaluation is actually correct but it is not an indeterminate form,$\log a-\log b=\log \frac{a}{b}$. So the limit is actually $\log 2$.

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    $\begingroup$ How about $\ln 2$? $\endgroup$ – Robert Lewis Mar 26 at 6:02
  • $\begingroup$ We usually mean the logarithm to the base of e when we say log with no base specified. $\endgroup$ – Boshu Mar 26 at 6:05
  • $\begingroup$ By we, I mean the way I was taught. I understand people do it differently elsewhere. $\endgroup$ – Boshu Mar 26 at 6:05
  • $\begingroup$ OK, then how about $\log 2$? I'm pointing out that the limit does not appear to be $2$! $\endgroup$ – Robert Lewis Mar 26 at 6:06
  • $\begingroup$ Oh I’m so sorry. I haven’t slept all night and I keep making these mistakes! Thanks for pointing it out. $\endgroup$ – Boshu Mar 26 at 6:09
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Another option is to substitute $u=t/x$ so your integral is $\int_1^2\frac{du}{u}$ regardless of $x$.

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Hence we got that

$$\int_x^{2x}\frac{\mathrm dt}t=\left[\log(t)\right]_x^{2x}=\log(2x)-\log(x)=\log(2)-\log(x)+\log(x)=\log(2)$$

Taking the limit as $x$ approaches infinity does not change the outcome, which is in fact independent of $x$. Thus

$$\lim_{x\to\infty}\int_x^{2x}\frac{\mathrm dt}t=\lim_{x\to\infty}\log(2)=\log(2)$$

$$\therefore~\lim_{x\to\infty}\int_x^{2x}\frac{\mathrm dt}t~=~\log2$$


Note that the independence of $x$ can be shown much easier. Write

$$\int_x^{2x}\frac{\mathrm dt}t=\int_x^{2x}\frac xt\frac{\mathrm dt}x\stackrel{\frac tx\mapsto t}=\int_1^2\frac{\mathrm dt}t=[\log t]_1^2=\log2$$

Taking the limit does not change the outcome as above.

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